Answer :

(a) for 10 mL of 0.2M Ca(OH)2 and 25 mL of 0.1M HCl

10 ml of 0.2 M Ca(OH)2= 10× 0.2 milimole = 2 milimoles


25 ml of 0.1 M HCl = 25× 0.1 milimole = 2.5 milimoles


Ca(OH)2 + 2HCl ↔ CaCl2 + 2H2O


0.1 a mole of Ca(OH)2 reacts with 2 millimole of HCl


∴ 2.5 mole of HCl reacts with 1.25 millimole of Ca(OH)2


Ca(OH)2 left = 1=1.25 = 0.75 milimole


Volume of reaction mixture = 10mL+ 25mL = 35mL


∴ Molarity of Ca(OH)2in the mixture solution is:


Molarity =


Hence,


Molarity = = 0.0214 M


⇒[OH-] = 2 × 0.0124 M


⇒[OH-] = 0.0428 M


To caculate pOH, we use the formula:


pH = - log[OH-]


⇒pH= - log (4.28× 10-2)


⇒pH = -log 4.28 + 2log 10


⇒pH = 2 – 0.6314


⇒pH = 1.37


As we know that, pH + pOH =14


∴ pH = 14 – 1.37


⇒pH = 12.63


Hence, the pH is 12.62


(b) for 10 mL of 0.01M H2SO4 and 10 mL of 0.01M Ca(OH)2


10 ml of 0.01 M H2sO4 = 10× 0.01 milimole = 0.1 milimole


10 ml of 0.01 M Ca(OH)2 = 10× 0.01 milimole = 0.1 milimole


Ca(OH)2 + H2sO4 = CaSO4 + 2H2O


1 mole of Ca(OH)2 reacts with 1 mole of H2sO4


∴ 0.1 millimole of Ca(OH)2 will react completely with 0.1 minimal of H2SO4. Hence solution will be neutral with pH= 7


(C) 10 mL of 0.1M H2SO4 and 10 mL of 0.1M KOH


10 ml of 0.01 M H2sO4 = 10× 0.01 milimole = 0.1 milimole


10 ml of 0.1 M KOH = 10× 0.1 milimole = 1 milimole


2KOH+ H2sO4 = K2SO4 + 2H2O


I mole of KOHreacts with 0.5 milimole of H2sO4


H2SO4 left = 1=0.5 = 0.5 milimole


Volume of reaction mixture = 10+ 10 = 20 mL


∴ Molarity of H2SO4 in the mixture solution is:


Molarity =


Hence,


Molarity = = 2.5 × 10-12 M


⇒[H+] = 2 × 2.5 × 10-12


⇒[H+] = 5× 10-2


To caculate pH, we use the formula:


pH = - log[H+]


⇒pH= - log (5× 10-2)


⇒pH = -log 5 + 2log 10


⇒pH = 2 – 0.699


⇒pH = 1.3


Hence, the pH is 1.3

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